Prove or disprove $\Bbb E[(X-\Bbb E[X \mid \mathcal F])^2\mid \mathcal F] = 0 \Rightarrow X $ is $\mathcal F$- measurable

Question

I have to prove or disprove for a random variable X with finite second moment that $$\Bbb E[(X-\Bbb E[X \mid \mathcal F])^2\mid \mathcal F] = 0 \Leftrightarrow X \:\mathcal F-\mathrm{ measurable}$$ I've shown the $\Leftarrow$ direction using that for an $\mathcal F$-measurable $X$ the following holds: $$\Bbb E[X\mid \mathcal F] = X$$ $$\mathrm{if } \: \Bbb E[|Y|], \Bbb E[|XY|] \lt \infty, \mathrm{then}\: \Bbb E[XY]=X\Bbb E[Y] $$ I'm now stuck at proving or disproving the other direction. Any tipps or ideas on how to do this?
Thanks in advance!

Answer 1

Taking the expectation on both sides, one see that $E[(X-E[X \mid \mathcal F])^2\mid \mathcal F] = 0$ implies $$ X=E[X \mid \mathcal F]\quad a.s. $$ and $E[X \mid \mathcal F]$ is $\mathcal{F}$-measurable.